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Mathematics 19 Online
OpenStudy (anonymous):

HEEELP ME !!!!!!! Determine the equation of each graph in the form y = acos(bx). https://www.virtualhighschool.com/content/OSSD-v2/Department/Math/courses/MCR3U/ver-D/Transforming_Trig/images/45.png?_&d2lSessionVal=zz5owIE4LYpM64Qh3q7GmzDHv and https://www.virtualhighschool.com/content/OSSD-v2/Department/Math/courses/MCR3U/ver-D/Transforming_Trig/images/46.png?_&d2lSessionVal=zz5owIE4LYpM64Qh3q7GmzDHv B) Determine the equation of each graph above in the form y = a sin(bx)

OpenStudy (anonymous):

PLEASE HELP!!

OpenStudy (anonymous):

in the first one it should be more or less clear that \(a=4\) because that is the highest point on the graph

OpenStudy (anonymous):

what is the equation then?

OpenStudy (anonymous):

\(b\) is found by setting \[\frac{360}{b}=1440\] and solving for \(b\) since the period of that function is \(1440\)

OpenStudy (anonymous):

i.e. \(b=\frac{360}{1440}=\frac{1}{4}\)

OpenStudy (anonymous):

wont b = 4?

OpenStudy (anonymous):

@tkhunny can you help me?

OpenStudy (anonymous):

b=4??

OpenStudy (anonymous):

no it will not be 4, it will be \(\frac{1}{4}\) for sure

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so the equation will be y= 4cos(1/4x) ?

OpenStudy (anonymous):

\[\frac{360}{b}=1440\iff 1440x=360\iff x=\frac{360}{1440}=\frac{1}{4}\]

OpenStudy (anonymous):

yes or \[4\cos(\frac{x}{4})\]

OpenStudy (anonymous):

can you do the second one??

OpenStudy (anonymous):

probably let me look

OpenStudy (anonymous):

this one looks like sine, but instead of starting at zero and going up, it starts at zero and goes down so \(a\) will be negative

OpenStudy (anonymous):

i can't really see how far down it goes what is the lowest point on the graph?

OpenStudy (anonymous):

-3

OpenStudy (anonymous):

ok so \(a=-3\)

OpenStudy (anonymous):

you do not see a full period here but you can figure it out from the picture, because half of the period is 90 degrees therefore the full period is 180 degrees

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

to find \(b\) as before, solve \[\frac{360}{b}=180\] for \(b\)

OpenStudy (anonymous):

the equation is -3cos(2x) ???

OpenStudy (anonymous):

not cosine, sine

OpenStudy (anonymous):

if it is right ...... can you do part b?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i thought that was B

OpenStudy (anonymous):

no you have to write the cos equation for the two graphs and then part b is writing the sine for the same two graphs

OpenStudy (anonymous):

@tkhunny help!!

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